1kaymant sir can u explain it plzzz??????
after quite sometime now!!!
well in this question i will give a question and 2 solutions to it...the people will have to tell which one is correct and why the other one is wrong!!!!
so here goes the question(IRODOV)::
A point positive charge q is placed in front of a flat surface at a perpendicular distance r from it! Now the charge is moved slowly to infinite separation from the plate. Find the work done in the process!
the solutions::
SOLUTION 1::
Consider an image charge -q at a distance r on the other side of the plane.
the potential energy content of the system is 14πεo.q22r
after the charge is taken to infinity potential energy of the system=0
so by conservation of energy,
work done = loss in potential energy = q28πεor
SOLUTION 2::
Let the separation of charge q from the surface be x. Consider an image charge -q at a distance x on the other side of the plate.
force of attraction on q charge=q216πεox2
on displacing it by distance dx, work done dW=q2dx16πεox2
thus, W=\int_{r}^{\propto}{\frac{q^2dx}{16 \pi \epsilon _ox^2} }=\frac{q^2}{16\pi\epsilon_o}\int_{r}^{\propto}{\frac{1}{x^2}}=\frac{q^2}{16\pi\epsilon_or}
the 2 processes give 2 different answers....thus invariably one is correct and the other is wrong....NOW TEMME WHCH PROCESS IS CORRECT AND WHY IS THE OTHER POCESS WRONG??
KALYAN a.k.a. Mr.Function and AVEEK a.k.a. Bicchuram are not allowed to POST IN THIS THREAD!! hehe!! ;)
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17 Answers
1Let ' s look at the behaviour of the phenomena caused by the charge " q " placed in front of the conducting plane . First , it will induce some charges on the plane itself . Then , there would arise a mutual attraction , due to which the charge " q " will gain some potential energy . But , if we try to move away the charge , then the charge distribution of the induced charges will change . In fact , the charge distribution on the plane at a distance " x " from the foot of the perpendicular drawn from the charge " q " when it is at a distance " d " from the plane could be derived as -
σ = - q d2 π ( d 2 + x 2 ) 3 / 2
Hence , if the charge distribution changes , then there would arise a field , which obviously will become time - dependant . So , it no longer will satisfy the concept of potential , as " Potential " could only exist in case of time invariant fields .
49haan...oops! sry forgot to mention dat!!
ye dil ki naadaniya maaf ho!! ;) LOL!!
106btw the question doesnt mention the word CONDUCTIONG PLATE at all [3]
1yaar can neone plzz explain me image method with an example plzzzzzzzzzz????????????
66Well here the concepts of potential field is still applicable. That's why the word "slowly" is stressed so that the field remains quasi static. The potential method would give the correct result provided you divided by 2. The reason for this factor of 2 becomes clear if one understands the basics of image method. The point is that we are considering two separate problems. One in which we have a point charge q at a distance d from a conducting plane. In the second case, we have actually two point charges. Only the halves of space (divided by the plane) in which the point charge q is present are electrically the same. So while considering the work by potential energy method, we must remove the contribution from the other half of the space where the negative charge is present.
49Hence , if the charge distribution changes , then there would arise a field , which obviously will become time - dependant . So , it no longer will satisfy the concept of potential , as " Potential " could only exist in case of time invariant fields .
Can u explain why??
thats correct! but now explain where the rest half of the existing energy is lost?? considering the fact that the work done is exactly half of the existing energy!!! :O
1The second process is correct . First of all , the charge placed will induce some charges on the plate , which in turn would create a field . But , the field caused by the induced charges is not a " Potential Field " . To make things sound more clear , it would be better if I say the integral of this field around a closed path is non - zero . That is why the concept of " Potential " fails here .
In other words , Work Done = Loss In Potential Energy
this relation will not be true in this case .
49Ricky has somewhat given the answer but his words are misleading so i wait till he reframes his answer and rectify it!!
49@ anirudh...u r wrong in ur concepts somewhere...
the basic assumption of image charge is based on that the force on the charge q due to the charge induced on the plate is equal to the force when we replace the plate charge -q on the specified position...
thus the energy content is same for the q-plate system and q-induced charge system!!
49Have u learnt Image Method???
The essence is that potential at the plate must be 0!!
potential of spherical shell = q/4πεor
and for flat surface, r=∞...
1i think first method is wrong because the energy content of the system is spread on both sides of the plane but in our actual question and the energy will be above the plane and not below. thus , potential energy will be half.
49@Ricky: look the induced charge attracts the q charge and that i why there is at all potential or potential energy....also hat is the precise reason that we need t do work in order to move the charge...
hope this clears ur doubt...
so thread still open....
Which is correct? and why?? [3]
